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Question Number 131037 by mohammad17 last updated on 31/Jan/21

Commented by mathmax by abdo last updated on 31/Jan/21

$$48)\mathrm{use}\mathrm{the}\mathrm{same}\mathrm{method}...$$

Answered by mathmax by abdo last updated on 31/Jan/21

$$\mathrm{I}={\int }_0^{\frac{1}{2}}{\mathrm{e}}^{-{\mathrm{x}}^3}\mathrm{dx}\Rightarrow \mathrm{I}={\int }_0^{\frac{1}{2}}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-{\mathrm{x}}^3)}^{\mathrm{n}}}{\mathrm{n}!}\mathrm{dx}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}!}{\int }_0^{\frac{1}{2}}{\mathrm{x}}^{3\mathrm{n}}\mathrm{dx}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}!}{\left[\frac{1}{3\mathrm{n}+1}{\mathrm{x}}^{3\mathrm{n}+1}\right]}_0^{\frac{1}{2}}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}!(3\mathrm{n}+1)2^{3\mathrm{n}+1}}$$$$\Rightarrow \mathrm{I}=\frac{1}{2}-\frac{1}{4.2^4}+\frac{1}{2!.7.2^7}-....$$$$\mathrm{you}\mathrm{can}\mathrm{use}5\mathrm{terms}\mathrm{of}\mathrm{this}\mathrm{serie}\mathrm{to}\mathrm{get}\mathrm{approximate}\mathrm{value}\mathrm{of}\mathrm{I}$$

Commented by mohammad17 last updated on 31/Jan/21

$$thankyousircanyouhelpmeinallquestionpleas$$

Answered by mathmax by abdo last updated on 31/Jan/21

$$\mathrm{J}={\int }_0^1\mathrm{xsin}\left({\mathrm{x}}^3\right)\mathrm{dx}\mathrm{we}\mathrm{have}\mathrm{sinu}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}{\mathrm{u}}^{2\mathrm{n}+1}}{(2\mathrm{n}+1)!}\mathrm{with}\mathrm{radius}\mathrm{R}=\mathrm{\infty }$$$$\Rightarrow \sin \left({\mathrm{x}}^3\right)=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{(2\mathrm{n}+1)!}{\mathrm{x}}^{6\mathrm{n}+3}\Rightarrow \mathrm{J}={\int }_0^1\mathrm{x}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{(2\mathrm{n}+1)!}{\mathrm{x}}^{6\mathrm{n}+3}\mathrm{dx}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{(2\mathrm{n}+1)!}{\int }_0^1{\mathrm{x}}^{6\mathrm{n}+4}\mathrm{dx}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{(2\mathrm{n}+1)!}{\left[\frac{1}{6\mathrm{n}+5}{\mathrm{x}}^{6\mathrm{n}+5}\right]}_0^1$$$$=\sum _{\mathrm{n}0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{(6\mathrm{n}+5)(2\mathrm{n}+1)!}\Rightarrow$$$$\mathrm{J}=\frac{1}{5}-\frac{1}{11.3!}+\frac{1}{17.5!}-\frac{1}{23.7!}+...$$$$5\mathrm{terms}\mathrm{give}\mathrm{a}\mathrm{best}\mathrm{approximation}\mathrm{of}\mathrm{J}$$$$$$

Commented by mohammad17 last updated on 31/Jan/21

$$nicesirthankyou$$

Commented by mathmax by abdo last updated on 31/Jan/21

$$\mathrm{another}\mathrm{we}\mathrm{know}\mathrm{u}-\frac{{\mathrm{u}}^3}{6}⩽\mathrm{sinu}⩽\mathrm{u}\Rightarrow {\mathrm{x}}^3-\frac{{\mathrm{x}}^9}{6}⩽\sin \left({\mathrm{x}}^3\right)⩽{\mathrm{x}}^3\Rightarrow$$$${\mathrm{x}}^4-\frac{{\mathrm{x}}^{10}}{6}⩽\mathrm{xsin}\left({\mathrm{x}}^3\right)⩽{\mathrm{x}}^4\Rightarrow {\int }_0^1({\mathrm{x}}^4-\frac{{\mathrm{x}}^{10}}{6})\mathrm{dx}⩽{\int }_0^1\mathrm{xsin}\left({\mathrm{x}}^3\right)\mathrm{dx}⩽{\int }_0^1{\mathrm{x}}^4\mathrm{dx}\Rightarrow$$$${[\frac{{\mathrm{x}}^5}{5}-\frac{1}{6.11}{\mathrm{x}}^{11}]}_0^1⩽\mathrm{J}⩽{\left[\frac{{\mathrm{x}}^5}{5}\right]}_0^1\Rightarrow \frac{1}{5}-\frac{1}{6\times 11}⩽\mathrm{J}⩽\frac{1}{5}$$$$\mathrm{so}{\mathrm{v}}_{\mathrm{o}}=\frac{1}{2}(\frac{1}{5}-\frac{1}{6.11}+\frac{1}{5})=\frac{1}{5}-\frac{1}{12.11}\mathrm{is}\mathrm{approximate}\mathrm{value}\mathrm{of}\mathrm{J}$$$$$$

Commented by mathmax by abdo last updated on 31/Jan/21

$$\mathrm{you}\mathrm{arewelcome}\mathrm{sir}.$$

Answered by mathmax by abdo last updated on 31/Jan/21

$$\mathrm{K}={\int }_0^{\frac{1}{2}}\frac{\mathrm{atctanx}}{\mathrm{x}}\mathrm{dx}\mathrm{we}\mathrm{have}\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{arctanx}\right)=\frac{1}{1+{\mathrm{x}}^2}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{x}}^{2\mathrm{n}}\Rightarrow$$$$\mathrm{arctanx}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{2\mathrm{n}+1}{\mathrm{x}}^{2\mathrm{n}+1}+\mathrm{c}(\mathrm{c}=0)=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{2\mathrm{n}+1}{\mathrm{x}}^{2\mathrm{n}+1}$$$$\mathrm{withradius}\mathrm{R}=1\Rightarrow \mathrm{K}={\int }_0^{\frac{1}{2}}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{2\mathrm{n}+1}{\mathrm{x}}^{2\mathrm{n}}\mathrm{dx}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{2\mathrm{n}+1}{\int }_0^{\frac{1}{2}}{\mathrm{x}}^{2\mathrm{n}}\mathrm{dx}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{2\mathrm{n}+1}{\left[\frac{1}{2\mathrm{n}+1}{\mathrm{x}}^{2\mathrm{n}+1}\right]}_0^{\frac{1}{2}}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{{(2\mathrm{n}+1)}^2.2^{2\mathrm{n}+1}}\Rightarrow$$$$\mathrm{K}=\frac{1}{2}-\frac{1}{3^2.2^3}+\frac{1}{5^2.2^5}-....$$

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